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Collapse of the Brazos River Bridge, Brazos, Texas, during erection of the 973-ft, continuous steel plategirders that support the roadway. The failure was initiated by overstress of the connections between theweb and flange during erection. Structures are particularly vulnerable to failure during erection becausestiffening elements—for example, floor slabs and bracing—may not be in place. In addition, the structure’sstrength may be reduced when certain connections are partially bolted or not fully welded to permit pre-cise alignment of members.

C H A P T E R

9Deflections of Beams and Frames

9.1 Introduction

When a structure is loaded, its stressed elements deform. In a truss, barsin tension elongate and bars in compression shorten. Beams bend andcables stretch. As these deformations occur, the structure changes shapeand points on the structure displace. Although these deflections are nor-mally small, as part of the total design, the engineer must verify that thesedeflections are within the limits specified by the governing design code toensure that the structure is serviceable. For example, large deflections ofbeams can lead to cracking of nonstructural elements such as plaster ceil-ings, tile walls, or brittle pipes. The lateral displacement of buildings pro-duced by wind forces must be limited to prevent cracking of walls andwindows. Since the magnitude of deflections is also a measure of a mem-ber’s stiffness, limiting deflections also ensures that excessive vibrationsof building floors and bridge decks are not created by moving loads.

Deflection computations are also an integral part of a number of ana-lytical procedures for analyzing indeterminate structures, computing buck-ling loads, and determining the natural periods of vibrating members.

In this chapter we consider several methods of computing deflectionsand slopes at points along the axis of beams and frames. These methodsare based on the differential equation of the elastic curve of a beam. Thisequation relates curvature at a point along the beam’s longitudinal axisto the bending moment at that point and the properties of the cross sec-tion and the material.

9.2 Double Integration Method

The double integration method is a procedure to establish the equationsfor slope and deflection at points along the longitudinal axis (elasticcurve) of a loaded beam. The equations are derived by integrating the

differential equation of the elastic curve twice, hence the name doubleintegration. The method assumes that all deformations are produced bymoment. Shear deformations, which are typically less than 1 percent ofthe flexural deformations in beams of normal proportions, are not usu-ally included. But if beams are deep, have thin webs, or are constructedof a material with a low modulus of rigidity (plywood, for example), themagnitude of the shear deformations can be significant and should beinvestigated.

To understand the principles on which the double integration methodis based, we first review the geometry of curves. Next, we derive the dif-ferential equation of the elastic curve—the equation that relates the cur-vature at a point on the elastic curve to the moment and the flexural stiff-ness of the cross section. In the final step we integrate the differentialequation of the elastic curve twice and then evaluate the constants ofintegration by considering the boundary conditions imposed by the sup-ports. The first integration produces the equation for slope; the secondintegration establishes the equation for deflection. Although the methodis not used extensively in practice since evaluating the constants of inte-gration is time-consuming for many types of beams, we begin our studyof deflections with this method because several other important proce-dures for computing deflections in beams and frames are based on thedifferential equation of the elastic curve.

Geometry of Shallow Curves

To establish the geometric relationships required to derive the differen-tial equation of the elastic curve, we will consider the deformations ofthe cantilever beam in Figure 9.1a. The deflected shape is represented inFigure 9.1b by the displaced position of the longitudinal axis (also calledthe elastic curve). As reference axes, we establish an x-y coordinate sys-tem whose origin is located at the fixed end. For clarity, vertical dis-tances in this figure are greatly exaggerated. Slopes, for example, aretypically very small—on the order of a few tenths of a degree. If we wereto show the deflected shape to scale, it would appear as a straight line.

To establish the geometry of a curved element, we will consider aninfinitesimal element of length ds located a distance x from the fixed end.As shown in Figure 9.1c, we denote the radius of the curved segment by r.At points A and B we draw tangent lines to the curve. The infinitesimalangle between these tangents is denoted by du. Since the tangents to thecurve are perpendicular to the radii at points A and B, it follows that theangle between the radii is also du. The slope of the curve at point Aequals

dy

dx� tan u

294 Chapter 9 Deflections of Beams and Frames

(a)P

(b)

y

x

x

A ds B

ds

dx

(c)

line tangent at B

line tangent at A

�

��A

B

d

�d

o

Figure 9.1

Section 9.2 Double Integration Method 295

If the angles are small (tan u � u radians), the slope can be written

(9.1)

From the geometry of the triangular segment ABo in Figure 9.1c, wecan write

(9.2)

Dividing each side of the equation above by ds and rearranging terms give

(9.3)

where du/ds, representing the change in slope per unit length of distancealong the curve, is called the curvature and denoted by the symbol c.Since slopes are small in actual beams, ds � dx, and we can express thecurvature in Equation 9.3 as

(9.4)

Differentiating both sides of Equation 9.1 with respect to x, we canexpress the curvature du/dx in Equation 9.4 in terms of rectangular coor-dinates as

(9.5)

Differential Equation of the Elastic Curve

To express the curvature of a beam at a particular point in terms of themoment acting at that point and the properties of the cross section, wewill consider the flexural deformations of the small beam segment of lengthdx, shown with darker shading in Figure 9.2a. The two vertical lines rep-resenting the sides of the element are perpendicular to the longitudinal axis

du

dx�

d 2y

dx 2

c �du

dx�

1r

c �du

ds�

1r

r du � ds

dy

dx� u

dx

x

(a)

d

N.A.

�

�

(b)

M

N.A.

(c)

�

( f )

d c

F

D

E

A

B �

d

dl

dx

�

�

M

(d)

�

�

(e)

cFigure 9.2: Flexural deformations of seg-ment dx: (a) unloaded beam; (b) loaded beamand moment curve; (c) cross section of beam;(d ) flexural deformations of the small beamsegment; (e) longitudinal strain; ( f ) flexuralstresses.

of the unloaded beam. As load is applied, moment is created, and thebeam bends (see Fig. 9.2b); the element deforms into a trapezoid as thesides of the segment, which remain straight, rotate about a horizontal axis(the neutral axis) passing through the centroid of the section (Fig. 9.2c).

In Figure 9.2d the deformed element is superimposed on the originalunstressed element of length dx. The left sides are aligned so that thedeformations are shown on the right. As shown in this figure, the longi-tudinal fibers of the segment located above the neutral axis shortenbecause they are stressed in compression. Below the neutral axis the lon-gitudinal fibers, stressed in tension, lengthen. Since the change in lengthof the longitudinal fibers (flexural deformations) is zero at the neutralaxis (N.A.), the strains and stresses at that level equal zero. The variationof longitudinal strain with depth is shown in Figure 9.2e. Since the strainis equal to the longitudinal deformations divided by the original lengthdx, it also varies linearly with distance from the neutral axis.

Considering triangle DFE in Figure 9.2d, we can express the changein length of the top fiber dl in terms of du and the distance c from theneutral axis to the top fiber as

(9.6)

By definition, the strain P at the top surface can be expressed as

(9.7)

Using Equation 9.6 to eliminate dl in Equation 9.7 gives

(9.8)

Using Equation 9.5 to express the curvature du/dx in rectangular coordi-nates, we can write Equation 9.8 as

(9.9)

If behavior is elastic, the flexural stress, s, can be related to the strainP at the top fiber by Hooke’s law, which states that

where E � the modulus of elasticity

Solving for P gives

(9.10)

Using Equation 9.10 to eliminate P in Equation 9.9 produces

(9.11)d 2y

dx 2 �s

Ec

P �s

E

s � EP

d 2y

dx 2 �Pc

P �du

dx c

P �dl

dx

dl � du c


Section 9.2 Double Integration Method 297

For elastic behavior the relationship between the flexural stress at the topfiber and the moment acting on the cross section is given by

(5.1)

Substituting the value of s given by Equation 5.1 into Equation 9.11 pro-duces the basic differential equation of the elastic curve

(9.12)

In Examples 9.1 and 9.2 we use Equation 9.12 to establish the equa-tions for both the slope and the deflection of the elastic curve of a beam.This operation is carried out by expressing the bending moment in termsof the applied load and distance x along the beam’s axis, substituting theequation for moment in Equation 9.12, and integrating twice. Themethod is simplest to apply when the loading and support conditions per-mit the moment to be expressed by a single equation that is valid over theentire length of the member—the case for Examples 9.1 and 9.2. Forbeams of constant cross section, E and I are constant along the length ofthe member. If E or I varies, it must also be expressed as a function of xin order to carry out the integration of Equation 9.12. If the loads or thecross section varies in a complex manner along the axis of the member,the equations for moment or for I may be difficult to integrate. For thissituation approximate procedures can be used to facilitate the solution(see, for example, the finite summation in Example 10.15).

For the cantilever beam in Figure 9.3a, establish the equations for slopeand deflection by the double integration method. Also determine the mag-nitude of the slope uB and deflection �B at the tip of the cantilever. EI isconstant.

SolutionEstablish a rectangular coordinate system with the origin at the fixedsupport A. Positive directions for the axes are up (y-axis) and to the right(x-axis). Since the slope is negative and becomes steeper in the positivex direction, the curvature is negative. Passing a section through the beama distance x from the origin and considering a free body to the right ofthe cut (see Fig. 9.3b), we can express the bending moment at the cut as

Substituting M into Equation 9.12 and adding a minus sign because thecurvature is negative lead to

d 2y

dx 2 �M

EI�

P 1L x 2EI

M � P 1L x 2

d 2y

dx 2 �M

EI

s �Mc

I

E X A M P L E 9 . 1

B

B

ox

A

L

yP

x

L – x

M

(a)

(b)

B

+dx–dy �

B�

PL – x

z

Figure 9.3[continues on next page]

Integrating twice to establish the equations for slope and deflection yields

(1)

(2)

To evaluate the constants of integration C1 and C2 in Equations 1 and 2,we use the boundary conditions imposed by the fixed support at A:

1. When x � 0, y � 0; then from Equation 2, C2 � 0.2. When x � 0, dy/dx � 0; then from Equation 1, C1 � 0.

The final equations are

(3)

(4)

To establish uB and �B, we substitute x � L in Equations 3 and 4 tocompute

Using the double integration method, establish the equations for slopeand deflection for the uniformly loaded beam in Figure 9.4. Evaluate thedeflection at midspan and the slope at support A. EI is constant.

SolutionEstablish a rectangular coordinate system with the origin at support A.Since the slope increases as x increases (the slope is negative at A, zeroat midspan, and positive at B), the curvature is positive. If we consider afree body of the beam cut by a vertical section located a distance x fromthe origin at A (see Fig. 9.4b), we can write the internal moment at thesection as

M �wLx

2

wx 2

2

¢B �PL3

3EI

uB �PL2

2EI

y �PLx 2

2EI

Px 3

6EI

u �dy

dx�

PLx

EI

Px 2

2EI

y �PLx 2

2EI

Px 3

6EI C1x C2

dy

dx�

PLx

EI

Px 2

2EI C1


E X A M P L E 9 . 2

Example 9.1 continues . . .

Section 9.3 Moment-Area Method 299

Substituting M into Equation 9.12 gives

(1)

Integrating twice with respect to x yields

(2)

(3)

To evaluate the constants of integration C1 and C2, we use the bound-ary conditions at supports A and B. At A, x � 0 and y � 0. Substitutingthese values into Equation 3, we find that C2 � 0. At B, x � L and y � 0.Substituting these values into Equation 3 and solving for C1 gives

Substituting C1 and C2 into Equations 2 and 3 and dividing both sides byEI yields

(4)

(5)

Compute the deflection at midspan by substituting x � L>2 intoEquation 5.

Compute the slope at A by substituting x � 0 into Equation 4.

9.3 Moment-Area Method

As we observed in the double integration method, based on Equation9.12, the slope and deflection of points along the elastic curve of a beamor a frame are functions of the bending moment M, moment of inertia I,

uA �dy

dx�

wL3

24EI

y �5wL4

384EI

y �wLx 3

12EI

wx 4

24EI

wL3x

24EI

u �dy

dx�

wLx 2

4EI

wx 3

6EI

wL3

24EI

C1 � wL3

24

0 �wL4

12

wL4

24 C1L

EIy �wLx 3

12

wx 4

24 C1x C2

EI dy

dx�

wLx 2

4

wx 3

6 C1

EI d 2y

dx 2 �wLx

2

wx 2

2A

A

B

o

(a)

w

x

R = wx

(b)

LwL2

y

+dx

+dx+dy–dy

w

wL2

wL2

M = – (wx)wLx2

x2

x2

x

Figure 9.4

and modulus of elasticity E. In the moment-area method we will estab-lish a procedure that utilizes the area of the moment diagrams [actually,the M>EI diagrams] to evaluate the slope or deflection at selected pointsalong the axis of a beam or frame.

This method, which requires an accurate sketch of the deflected shape,employs two theorems. One theorem is used to calculate a change inslope between two points on the elastic curve. The other theorem is usedto compute the vertical distance (called a tangential deviation) betweena point on the elastic curve and a line tangent to the elastic curve at a sec-ond point. These quantities are illustrated in Figure 9.5. At points A andB, tangent lines, which make a slope of uA and uB with the horizontal axis,are drawn to the elastic curve. For the coordinate system shown, the slopeat A is negative and the slope at B is positive. The change in slope betweenpoints A and B is denoted by �uAB. The tangential deviation at point B—the vertical distance between point B on the elastic curve and point C onthe line drawn tangent to the elastic curve at A—is denoted as tBA. Wewill use two subscripts to label all tangential deviations. The first sub-script indicates the location of the tangential deviation; the second sub-script specifies the point at which the tangent line is drawn. As you cansee in Figure 9.5, tBA is not the deflection of point B (vB is the deflection).With some guidance you will quickly learn to use tangential deviationsand changes in slope to compute values of slope and deflection at anydesired point on the elastic curve. In the next section we develop the twomoment-area theorems and illustrate their application to a variety of beamsand frames.

Derivation of the Moment-Area Theorems

Figure 9.6b shows a portion of the elastic curve of a loaded beam. Atpoints A and B tangent lines are drawn to the curve. The total anglebetween the two tangents is denoted by �uAB. To express �uAB in termsof the properties of the cross section and the moment produced by theapplied loads, we will consider the increment of angle change du thatoccurs over the length ds of the infinitesimal segment located a distancex to the left of point B. Previously, we established that the curvature at apoint on the elastic curve can be expressed as

(9.12)

where E is the modulus of elasticity and I is the moment of inertia. Mul-tiplying both sides of Equation 9.12 by dx gives

(9.13)du �M

EI dx

du

dx�

M

EI


x

A B

BBAt

�

�A�AB��

undeflectedbeam

tangentat B

tangentat A

elasticcurve

B

C

y

Figure 9.5


To establish the total angle change �uAB, we must sum up the du incre-ments for all segments of length ds between points A and B by integration.

(9.14)

We can evaluate the quantity M dx>EI in the integral of Equation 9.14graphically by dividing the ordinates of the moment curve by EI to pro-duce an M>EI curve (see Fig. 9.6c). If EI is constant along the beam’saxis the (most common case), the M>EI curve has the same shape as themoment diagram. Recognizing that the quantity M dx>EI represents aninfinitesimal area of height M>EI and length dx (see the crosshatchedarea in Fig. 9.6c), we can interpret the integral in Equation 9.14 as rep-resenting the area under the M>EI diagram between points A and B. Thisrelationship constitutes the first moment-area principle, which can bestated as

The change in slope between any two points on a continuous elasticcurve is equal to the area under the M>EI curve between these points.

You will notice that the first moment-area theorem applies only to thecase where the elastic curve is continuous between two points. If a hingeoccurs between two points, the area under the M>EI diagram will notaccount for the difference in slope that can exist on either side of thehinge. Therefore, we must determine the slopes at a hinge by workingwith the elastic curve on either side.

¢uAB � �B

A

du � �B

A

M dx

EI

�

(a) Moment diagram

(c)

(b)

tangent at A

diagram

tangent at Belasticcurve

EI = constant

A

MA MB

AB��

1 2

d�

d� tBA

MA

A B

EI

MB

EIMEI

MEI

dtdx

x

x

dx

ds

B

w

Figure 9.6

To establish the second moment-area theorem, which enables us toevaluate a tangential deviation, we must sum the infinitesimal incrementsof length dt that make up the total tangential deviation tBA (see Fig. 9.6b).The magnitude of a typical increment dt contributed to the tangentialdeviation tBA by the curvature of a typical segment ds between points 1and 2 on the elastic curve can be expressed in terms of the angle betweenthe lines tangent to the ends of the segment and the distance x betweenthe segment and point B as

(9.15)

Expressing du in Equation 9.15 by Equation 9.13, we can write

(9.16)

To evaluate tBA, we must sum all increments of dt by integrating the con-tribution of all the infinitesimal segments between points A and B:

(9.17)

Remembering that the quantity M dx>EI represents an infinitesimal areaunder the M>EI diagram and that x is the distance from that area to pointB, we can interpret the integral in Equation 9.17 as the moment aboutpoint B of the area under the M>EI diagram between points A and B. Thisresult constitutes the second moment-area theorem, which can be statedas follows:

The tangential deviation at a point B on a continuous elastic curve fromthe tangent line drawn to the elastic curve at a second point A is equalto the moment about B of the area under the M>EI diagram between thetwo points.

Although it is possible to evaluate the integral in Equation 9.17 byexpressing the moment M as a function of x and integrating, it is fasterand simpler to carry out the computation graphically. In this procedurewe divide the area of the M>EI diagram into simple geometric shapes—rectangles, triangles, parabolas, and so forth. Then the moment of eacharea is evaluated by multiplying each area by the distance from its cen-troid to the point at which the tangential deviation is to be computed. Forthis computation, we can use Table A.1 (see Appendix), which tabulatesproperties of areas you will frequently encounter.

tBA � �B

A

dt � �B

A

Mx

EI dx

dt �M dx

EI x

dt � du x



Application of the Moment-Area Theorems

The first step in computing the slope or deflection of a point on the elas-tic curve of a member is to draw an accurate sketch of the deflectedshape. As discussed in Section 5.6, the curvature of the elastic curve mustbe consistent with the moment curve, and the ends of members must sat-isfy the constraints imposed by the supports. Once you have constructeda sketch of the deflected shape, the next step is to find a point on the elas-tic curve where the slope of a tangent to the curve is known. After thisreference tangent is established, the slope or deflection at any other pointon the elastic curve can easily be established by using the moment-areatheorems.

The strategy for computing slopes and deflections by the moment-areamethod will depend on how a structure is supported and loaded. Most con-tinuous members will fall into one of the following three categories:

1. Cantilevers2. Structures with a vertical axis of symmetry that are loaded

symmetrically3. Structures that contain a member whose ends do not displace in the

direction normal to the original position of the member’slongitudinal axis

If a member is not continuous because of a hinge, the deflection atthe hinge must be computed initially to establish the position of the end-points of the member. This procedure is illustrated in Example 9.10. Inthe next sections we discuss the procedure for computing slopes anddeflections for members in each of the foregoing categories.

Case 1. In a cantilever, a tangent line of known slope can be drawn tothe elastic curve at the fixed support. For example, in Figure 9.7a the linetangent to the elastic curve at the fixed support is horizontal (i.e., theslope of the elastic curve at A is zero because the fixed support preventsthe end of the member from rotating). The slope at a second point B onthe elastic curve can then be computed by adding algebraically, to theslope at A, the change in slope �uAB between the two points. This rela-tionship can be stated as

(9.18)

where uA is the slope at the fixed end (that is, uA � 0) and �uAB is equalto the area under the M>EI diagram between points A and B.

Since the reference tangent is horizontal, tangential deviations—thevertical distance between the tangent line and the elastic curve—are, infact, displacements. Examples 9.3 to 9.5 cover the computation of slopes

uB � uA ¢uAB

AB

P

A = 0�

(a)

referencetangent

Figure 9.7: Position of tangent line: (a) can-tilever, point of tangency at fixed support.

and deflections in cantilevers. Example 9.4 illustrates how to modify anM>EI curve for a member whose moment of inertia varies. In Example9.5 the moment curves produced by both a uniform and a concentratedload are plotted separately in order to produce moment curves with aknown geometry. (See Table A.1 for the properties of these areas.)

Case 2. Figures 9.7b and c show examples of symmetric structuresloaded symmetrically with respect to the vertical axis of symmetry at thecenter of the structure. Because of symmetry the slope of the elastic curveis zero at the point where the axis of symmetry intersects the elasticcurve. At this point the tangent to the elastic curve is horizontal. For thebeams in Figure 9.7b and c we conclude, based on the first moment-areaprinciple, that the slope at any point on the elastic curve equals the areaunder the M>EI curve between that point and the axis of symmetry.

The computation of deflections for points along the axis of the beamin Figure 9.7c, which has an even number of spans, is similar to that ofthe cantilever in Figure 9.7a. At the point of tangency (point B), both thedeflection and slope of the elastic curve equal zero. Since the tangent tothe elastic curve is horizontal, deflections at any other point are equal totangential deviations from the tangent line drawn to the elastic curve atsupport B.

When a symmetric structure consists of an odd number of spans (one,three, and so on), the foregoing procedure must be modified slightly. For


P P P

A A

B

C

B D

E

tBC

tBA

Cv

tCA tBA

Cv

P

P

C = 0�

A�A�

(b)

C

referencetangent

referencetangent

La a

(d)

� �

(c)

L L

(e)

referencetangent

L

C�

PA B

referencetangent

B = 0�

A BCx

Figure 9.7: (b) and (c) symmetric members withsymmetric loading, point of tangency at intersec-tion of axis of symmetry and elastic curve; and(d ) and (e) point of tangency at left end of mem-ber AB.


example, in Figure 9.7b we observe that the tangent to the elastic curveis horizontal at the axis of symmetry. Computation of slopes will againbe referenced from the point of tangency at C. However, the centerline ofthe beam has displaced upward a distance vC; therefore, tangential devi-ations from the reference tangents are usually not deflections. We cancompute vC by noting that the vertical distance between the tangent lineand the elastic curve at either support B or C is a tangential deviation thatequals vC. For example, in Figure 9.7b vC equals tBC. After vC is com-puted, the deflection of any other point that lies above the original posi-tion of the unloaded member equals vC minus the tangential deviation ofthe point from the reference tangent. If a point lies below the undeflectedposition of the beam (for example, the tips of the cantilever at A or E),the deflection is equal to the tangential deviation of the point minus vC.Examples 9.6 and 9.7 illustrate the computation of deflections in a sym-metric structure.

Case 3. The structure is not symmetric but contains a member whoseends do not displace in a direction normal to the member’s longitudinalaxis. Examples of this case are shown in Figure 9.7d and e. Since theframe in Figure 9.7d is not symmetric and the beam in Figure 9.7e is notsymmetrically loaded, the point at which a tangent to the elastic curve ishorizontal is not initially known. Therefore, we must use a sloping tan-gent line as a reference for computing both slopes and deflections atpoints along the elastic curve. For this case we establish the slope of theelastic curve at either end of the member. At one end of the member, wedraw a tangent to the curve and compute the tangential deviation at theopposite end. For example, in either Figure 9.7d or e, because deflectionsare small the slope of the tangent to the elastic curve at A can be written

(9.19)

Since tan uA � uA in radians, we can write Equation 9.19 as

At a second point C, the slope would equal

where �uAC equals the area under the M>EI curve between points A and C.To compute the displacements of a point C located a distance x to the

right of support A (see Fig. 9.7e), we first compute the vertical distanceCC� between the initial position of the longitudinal axis and the refer-ence tangent. Since uA is small, we can write

CC¿ � uA 1x 2

uC � uA ¢uAC

uA �tBA

L

tan uA �tBA

L

The difference between CC� and the tangential deviation tCA equals thedeflection vC:

Examples 9.8 to 9.12 illustrate the procedure to compute slopes anddeflections in members with inclined reference tangents.

If the M>EI curve between two points on the elastic curve containsboth positive and negative areas, the net angle change in slope betweenthose points equals the algebraic sum of the areas. If an accurate sketchof the deflected shape is drawn, the direction of both the angle changesand the deflections are generally apparent, and the student does not haveto be concerned with establishing a formal sign convention to establishif a slope or deflection increases or decreases. Where the moment is pos-itive (see Fig. 9.8a), the member bends concave upward, and a tangentdrawn to either end of the elastic curve will lie below the curve. In otherwords, we can interpret a positive value of tangential deviation as anindication that we move upward from the tangent line to the elasticcurve. Conversely, if the tangential deviation is associated with a nega-tive area under the M>EI curve, the tangent line lies above the elasticcurve (see Fig. 9.8b), and we move downward vertically from the tangentline to reach the elastic curve.

Compute the slope uB and the deflection vB at the tip of the cantileverbeam in Figure 9.9a. EI is constant.

SolutionDraw the moment curve and divide all ordinates by EI (Fig. 9.9b).

Compute uB by adding to the slope at A the change in slope �uAB

between points A and B. Since the fixed support prevents rotation, uA � 0.

(1)

By the first moment-area theorem, �uAB equals the area under the trian-gular M>EI curve between points A and B.

(2)

Substituting Equation 2 into Equation 1 gives

uB � PL2

2EI

¢uAB �1

21L 2 aPL

EIb �

PL2

2EI

uB � uA ¢uAB � ¢uAB

vC � CC¿ tCA


+M+M

–M–M

AB

AB

tBA

tBA

tangent at A

tangent at A

(a)

(b)

Figure 9.8: Position of reference tangent: (a) pos-itive moment; (b) negative moment.

E X A M P L E 9 . 3

= 0P

P

BA

A�

B�

= 0A

= tBAB

M = PL

(a)

(b)

– PLEI

A Bx = L

23

L

MEI

curve

v

Figure 9.9


Since the tangent line at B slopes downward to the right, its slope is neg-ative. In this case the negative ordinate of the M>EI curve gave the cor-rect sign. In most problems the direction of the slope is evident from thesketch of the deflected shape.

Compute the deflection vB at the tip of the cantilever using the sec-ond moment-area theorem. The black dot in the M>EI curve denotes thecentroid of the area.

Beam with a Variable Moment of Inertia

Compute the deflection of point C at the tip of the cantilever beam in Fig-ure 9.10 if E � 29,000 kips/in2, IAB � 2I, and IBC � I, where I � 400 in4.

SolutionTo produce the M>EI curve, the ordinates of the moment curve are dividedby the respective moments of inertia. Since IAB is twice as large as IBC,the ordinates of the M>EI curve between A and B will be one-half the sizeof those between B and C. Since the deflection at C, denoted by vC,equals tCA, we compute the moment of the area of the M>EI diagramabout point C. For this computation, we divide the M>EI diagram intotwo rectangular areas.

where 1728 converts cubic feet to cubic inches.

vC �4500 11728 229,000 1400 2 � 0.67 in

vC � tCA �100

2EI 16 2 19 2

100

EI 16 2 13 2 �

4500

EI

vB �1

2 L aPL

EIb 2L

3�

PL3

3EI 1minus sign indicates that the

tangent line lies above elastic curve 2

vB � tBA � moment of triangular area ofM>EI diagram about point B

E X A M P L E 9 . 4

A

+M

CtCA = c

C�

M = 100 kip •ft

xM curve

diagram

100 kip •ft

B I2I

(a)

(b)

(c)

50EI

6� 6�

3�

9�

MEI

100EI

Figure 9.10: (a) Deflected shape; (b) momentcurve; (c) M>EI diagram divided into two rectan-gular areas.

Use of Moment Curve by “Parts”

Compute the slope of the elastic curve at B and C and the deflection at Cfor the cantilever beam in Figure 9.11a; EI is constant.

SolutionTo produce simple geometric shapes in which the location of the centroidis known, the moment curves produced by the concentrated load P and theuniform load w are plotted separately and divided by EI in Figure 9.11band c. Table A.1 provides equations for evaluating the areas of commongeometric shapes and the position of their centroids.

Compute the slope at C where �uAC is given by the sum of the areasunder the M>EI diagrams in Figure 9.11b and c; uA � 0 (see Fig. 9.11d).

Compute the slope at B. The area between A and B in Figure 9.11c iscomputed by deducting the parabolic area between B and C in Figure9.11c from the total area between A and C. Since the slope at B is smallerthan the slope at C, the area between B and C will be treated as a posi-tive quantity to reduce the negative slope at C.

Compute �C, the deflection at C. The deflection at C equals the tangen-tial deviation of C from the tangent to the elastic curve at A (see Fig. 9.11d).

� 4032

EI

� 1

2 16 2 a48

EIb 16 4 2

1

3 112 2 a72

EIb 19 2

¢C � tCA � moments of areas under M>EI curves betweenA and C in Figure 9.11b and c

� 396

EI radians

� 432

EI

1

316 2 a 18

EIb

uB � uC ¢uBC

� 432

EI radians

� 0 1

216 2 a48

EIb

1

3112 2 a72

EIb

uC � uA ¢uAC


E X A M P L E 9 . 5

A B

P = 8 kips

w = 1 kip/ft

C

(a)

(b)

C

C

C

(c)

tangent at A

(d)

6�

x = 10�

x = 9�

�c = tCA

�C

�A = 0

6�

–48EI

–72EI

–18EI

MEI

MEI

6�

6�

�B

Figure 9.11: Moment curve by “parts”: (a) beam;(b) M>EI curve associated with P; (c) M>EI curveassociated with uniform load w; (d) deflectedshape.


Analysis of a Symmetric Beam

For the beam in Figure 9.12a, compute the slope at B and the deflectionsat midspan and at point A. Also EI is constant.

SolutionBecause both the beam and its loading are symmetric with respect to thevertical axis of symmetry at midspan, the slope of the elastic curve iszero at midspan and the tangent line at that point is horizontal. Since nobending moments develop in the cantilevers (they are unloaded), the elas-tic curve is a straight line between points A and B and points D and E.See Appendix for geometric properties of a parabolic area.

Compute uB.

�wL3

24EI

� 0 2

3a L

2b awL2

8EIb

uB � uC ¢uCB

E X A M P L E 9 . 6

A EB

axis of symmetry

C

w

D

B�B�

A

C

(b)

(c)

straight

tangent at C

straight

A

B CtBC

(a)

L2

L2

L3

wL2

8EI5L16

x =

MEI

L3

vv

Figure 9.12: (a) Symmetric beam; (b) M>EI dia-gram; (c) geometry of the deflected shape.

[continues on next page]

Compute vC. Since the tangent at C is horizontal, vC equals tBC. Usingthe second moment-area theorem, we compute the moment of the para-bolic area between B and C about B.

Compute vA. Since the cantilever AB is straight,

where uB is evaluated in the first computation.

The beam in Figure 9.13a supports a concentrated load P at midspan(point C). Compute the deflections at points B and C. Also compute theslope at A. EI is constant.

SolutionCompute uA. Since the structure is symmetrically loaded, the slope of theline tangent to the elastic curve at midspan is zero; that is, uC � 0 (seeFig. 9.13c).

where �uAC is equal to the area under M>EI curve between A and C.

Compute vC, the deflection at midspan. Since the tangent at C is hor-izontal, vC � tAC, where tAC equals the moment about A of the triangulararea under the M>EI curve between A and C.

(1)

Compute vB, the deflection at the quarter point. As shown in Figure9.13c,

(2)

where tBC is the moment about B of the area under the M>EI curvebetween B and C. For convenience, we divide this area into a triangle anda rectangle. See the shaded area in Figure 9.13b.

tBC �1

2 a L

4b a PL

8EIb a L

6b

L

4 a PL

8EIb a L

8b �

5PL3

768EI

vB tBC � vC �PL3

48EI

vC �1

2a L

2b a PL

4EIb a 2

3 L

2b �

PL3

48EI

uA � 0 1

2 a L

2b a PL

4EIb �

PL2

16EI radians

uA � uC ¢uAC

vA � uB L

3�

wL3

24EI L

3�

wL4

72EI

vC � tBC �2

3a L

2b awL2

8EIb a 5L

16b �

5wL4

384EI


E X A M P L E 9 . 7

B C

A�

C� = 0tangent at C

(a)

(b)

(c)

L2

L4

L4

PL4EI

MEI

PL8EI

PL8EI

A

A B C D

tACtBC

B C D

P

L6

x =

L8

x =

Figure 9.13: (a) Beam details; (b) M>EI curve;(c) deflected shape.



Substituting tBC into Equation 2, we compute vB.

Analysis Using a Sloping Reference Tangent

For the steel beam in Figure 9.14a, compute the slope at A and C. Alsodetermine the location and value of the maximum deflection. If the max-imum deflection is not to exceed 0.6 in, what is the minimum requiredvalue of I? We know that EI is constant and E � 29,000 kips/in2.

SolutionCompute the slope uA at support A by drawing a line tangent to the elas-tic curve at that point. This will establish a reference line of known direc-tion (see Fig. 9.14c).

(1)

Since for small angles tan uA � uA (radians), Equation 1 can be written

(2)

where the expression for the moment arm is given in the right-hand col-umn of Table A.1, case (a). Substituting tCA into Equation 2 gives

A minus sign is added, because moving in the positive x direction, the tan-gent line, directed downward, has a negative slope.

Compute uC.

where �uAC equals area under M>EI curve between A and C.

uC � 384

EI

1

2118 2 a 96

EIb �

480

EI radians

uC � uA ¢uAC

uA �6912>EI

18�

384

EI radians

� 1

2 118 2 a 96

EIb a 18 6

3b �

6912

EI

tCA � moment of M>EI area between A and C about C

uA �tCA

L

tan uA �tCA

L

vB �11PL3

768EI

RA = 8 kips RC = 16 kips

P = 24 kips

A�

D� = 0D

C�

(a)

96EI

(b)

(c)

12� 6�

A B C

x

A

tangent at A

D

E

tDA tCA

C

f

g

e

da

y

Figure 9.14: (a) Beam; (b) M>EI diagram;(c) geometry of deflected shape.

E X A M P L E 9 . 8


Compute the maximum deflection. The point of maximum deflectionoccurs at point D where the slope of the elastic curve equals zero (thatis, uD � 0). To determine this point, located an unknown distance x fromsupport A, we must determine the area under the M>EI curve between Aand D that equals the slope at A. Letting y equal the ordinate of the M>EIcurve at D (Fig. 9.14b) gives

(3)

Expressing y in terms of x by using similar triangles afg and aed (see Fig.9.14b) yields

(4)

Substituting the foregoing value of y into Equation 3 and solving for x give

Substituting x into Equation 4 gives

Compute the maximum deflection vD at x � 9.8 ft

(5)

where the terms in Equation 5 are illustrated in Figure 9.14c.

Substituting DE and tDA into Equation 5 gives

(6)

Compute Imin if vD is not to exceed 0.6 in; in Equation 6 set vD � 0.6 inand solve for Imin.

Imin � 249.1 in4

vD �2508.3 11728 2

29,000Imin� 0.6 in

vD �3763.2

EI

1254.9

EI�

2508.3

EI

tDA � 1areaAD 2 x �1

2 19.8 2 a 78.4

EIb a 9.8

3b �

1254.9

EI

DE � uA# x �

384

EI 19.8 2 �

3763.2

EI

vD � DE tDA

y �78.4

EI

x � 9.8 ft

y �8x

EI

96> 1EI 212

�yx

0 � 384

EI

1

2 xy

uD � uA ¢uAD




For the beam in Figure 9.15, compute the slope of the elastic curve at pointsA and C. Also determine the deflection at A. Assume rocker at C equiv-alent to a roller.

SolutionSince the moment curve is negative at all sections along the axis of thebeam, it is bent concave downward (see the dashed line in Fig. 9.15c). Tocompute uC, we draw a tangent to the elastic curve at point C and com-pute tBC.

where

(Since the tangent line slopes downward to the right, the slope uC isnegative.)

Compute uA.

where �uAC is the area under the M>EI curve between A and C. Since theelastic curve is concave downward between points A and C, the slope atA must be opposite in sense to the slope at C; therefore, �uAC must betreated as a positive quantity.

Compute dA.

where

[See case (a) in Table A.1 for the equation for .]

Y � 24uC � 24 a 540

EIb �

12,960

EI

x

tAC � areaAC# x �

1

2 124 2 a 180

EIb a 6 24

3b �

21,600

EI

dA � tAC Y 1see Fig. 9.15c 2 �8640

EI

uA � 540

EI

1

2 124 2 a 180

EIb �

1620

EI

uA � uC ¢uAC

tBC � areaBC# x �

1

2118 2 a

180

EIb a 18

3b �

9720

EI

uC �tBC

18�

9720

EI a 1

18b �

540

EI

E X A M P L E 9 . 9

P = 30 kips

A�

C�

A�

(a)

rocker

(b)

(c)

– 180EI

10 kips40 kips

6� 18�

MEI

tAC tBCY

B C

tangent at C

B CA

Figure 9.15: (a) Beam, (b) M>EI diagram,(c) geometry of deflected shape.

The beam in Figure 9.16a contains a hinge at B. Compute the deflection vB

of the hinge, the slope of the elastic curve at support E, and the end slopesuBL and uBR of the beams on either side of the hinge (see Fig. 9.16d). Alsolocate the point of maximum deflection in span BE. EI is constant. The elas-tomeric pad at E is equivalent to a roller.

SolutionThe deflection of the hinge at B, denoted by vB, equals tBA, the tangentialdeviation of B from the tangent to the fixed support at A. Deflection tBA

equals the moment of the area under the M>EI curve between A and Babout B (see Fig. 9.16b).

vB � tBA � area # x �1

2 a

108

EIb 19 2 16 2 �

2916

EI


E X A M P L E 9 . 1 0

A� = 0

B = tBA

F�

E�

BL�

BR�= 0

(a)

(b)

(d)

(c)

–

72EI

tangent at E

tangent at AA B

tBE

x

E

F

B

hingeelastomeric

pad

A

RE = 12 kips

B C D E

12 kips108 kip •ft

12 kips

12 kips

MEI

108EI

x = 6�

9� 6� 6� 6�

Figure 9.16: (a) Beam with hinge at B;(b) deflected shape; (c) M>EI curve; (d ) detailshowing the difference in slope of the elasticcurve on each side of the hinge.


Compute uBL, the slope of the B end of cantilever AB.

where �uAB is equal to the triangular area under the M>EI curve betweenA and B and uA � 0 because the fixed support at A prevents rotation.

Compute uE, the slope of the elastic curve at E (see Fig. 9.16b).

where tBE equals the moment of the area under the M>EI curve between Band E about B. This computation is simplified by dividing the trapezoidalarea into two triangles and a rectangle (see the dashed lines in Fig. 9.16c).

Locate the point of maximum deflection in span BE. The point of max-imum deflection, labeled point F, is located at the point in span BE wherethe tangent to the elastic curve is zero. Between F and support E, a distancex, the slope goes from 0 to uE. Since the change in slope is given by thearea under the M>EI curve between these two points, we can write

(1)

where uF � 0 and uE � 594>EI rad. Between points D and E the changein slope produced by the area under the M>EI curve equals 216>EI. Sincethis value is less than uE, the slope at D has a positive value of

(2)

Between D and C the area under the M>EI curve equals 432>EI. Since thisvalue of change in slope exceeds 378>EI, the point of zero slope must liebetween C and D. We can now use Equation 1 to solve for distance x.

Compute uBR.

� 270

EI radians

�594

EI c 72

EI 16 2

1

2 16 2 a 72

EIb 12 2 d

uBR � uE ¢uBE

x � 11.25 ft

594

EI� 0

1

2a 72

EIb 16 2

72

EI 1x 6 2

uD � uE ¢uED �594

EI

216

EI�

378

EI radians

uE � uF ¢uEF

tBE �1

216 2 a72

EIb 14 2 16 2 a72

EIb 19 2 1

216 2 a72

EIb 114 2 � 7776

EI radians

uE �vB tBE

18� a 2916

EI

7776

EIb a 1

18b �

594

EI radians

� 0 1

2 19 2 a108

EIb �

486

EI rad

uBL � uA ¢uAB

Determine the deflection of the hinge at C and the rotation of joint B forthe braced frame in Figure 9.17a. For all members EI is constant.

SolutionTo establish the angular rotation of joint B, we consider the deflectedshape of member AB in Figure 9.17b. (Because member BCD contains ahinge, its elastic curve is not continuous, and it is not possible initially tocompute the slope at any point along its axis.)

Deflection of hinge:

� 16 2 a 288

EIb

1

216 2 a 72

EIb 14 2 �

2592

EI

¢ � 6uB tCB

uB �tAB

12�

1

2 12

72

EI 18 2

12�

288

EI


E X A M P L E 9 . 1 1

B C D

tCB

A

B D

tAB

A

12 kips

6 kips

12 kips

hinge

24 kips

6 kips

90° 6

24 kips

B�

B�

B�

(a)

(b)6� 6�6�

72EI

72EI

72EI

x = 8�

x = 4�

12�

Figure 9.17: (a) Frame and M>EI curves;(b) deflected shape.


Compute the horizontal deflection of joint B of the frame shown in Fig-ure 9.18a. EI is constant all members. Assume elastomeric pad at C actsas a roller.

SolutionBegin by establishing the slope of the girder at joint B.

(1)uB �tCB

L

E X A M P L E 9 . 1 2

12�

12

tangent at B

vertical line

tangent at B

B�B�

B C�C

C

A

B

A D

10 kips

20 kips

P = 20 kips 10 kips

(a)

12�

120EI

120EI

x = 9�

x = 8�

x = 4�

6�rigid joint

elastomericpad

6�

tCB

tAB

B� B�B�

B�

B�

B�

B�

(b)

(c)

deflectedposition

90°

Figure 9.18: (a) Frame and M>EI curves; (b) de-flected shape; (c) detail of joint B in deflectedposition.


where

Thus

Because joint B is rigid, the top of column AB also rotates through anangle uB (see Fig. 9.18c). Since the deflection �B at joint B is equal to thehorizontal distance AD at the base of the column, we can write

where tAB equals the moment of the M>EI diagram between A and Babout A, and the M>EI diagram is broken into two areas.

9.4 Elastic Load Method

The elastic load method is a procedure for computing slopes and deflec-tions in simply supported beams. Although the calculations in this methodare identical to those of the moment-area method, the procedure appearssimpler because we replace computations of tangential deviations andchanges in slope with the more familiar procedure of constructing shearand moment curves for a beam. Thus the elastic load method eliminatesthe need (1) to draw an accurate sketch of the member’s deflected shapeand (2) to consider which tangential deviations and angle changes to eval-uate in order to establish the deflection or the slope at a specific point.

In the elastic load method, we imagine that the M>EI diagram, whoseordinates represent angle change per unit length, is applied to the beam asa load (the elastic load). We then compute the shear and moment curves. Aswe will demonstrate next, the ordinates of the shear and the moment curvesat each point equal the slope and deflection, respectively, in the real beam.

To illustrate that the shear and moment at a section produced by anangle change, applied to a simply supported beam as a fictitious load,equal the slope and deflection at the same section, we examine the deflectedshape of a beam whose longitudinal axis is composed of two straightsegments that intersect at a small angle u. The geometry of the bentmember is shown by the solid line in Figure 9.19.

� 13,680

EI

� 120

EI 16 2 19 2

1

2 a 120

EIb 16 2 14 2 112 2 a 480

EIb

¢B � AD � tAB 12uB

uB �5760

EI a 1

12b �

480

EI radians

tCB �1

2a 120

EIb 112 2 18 2 �

5760

EI and L � 12 ft


A�

�A�

L

x

B CA

B�

C�

L – x

�C

�B

Figure 9.19: Beam with an angle change of u atpoint B.


Section 9.4 Elastic Load Method 319

If the beam ABC� is connected to the support at A so that segment ABis horizontal, the right end of the beam at C� will be located a distance�C above support C. In terms of the dimensions of the beam and theangle u (see triangle C�BC), we find

(1)

The sloping line AC�, which connects the ends of the beam, makes anangle uA with a horizontal axis through A. Considering the right triangleACC�, we can express uA in terms of �C as

(2)

Substituting Equation 1 into Equation 2 leads to

(3)

We now rotate member ABC� clockwise about the pin at A until chordAC� coincides with the horizontal line AC and point C� rests on the rollerat C. The final position of the beam is shown by the heavy dashed lineAB�C. As a result of the rotation, segment AB slopes downward to theright at an angle uA.

To express �B, the vertical deflection at B, in terms of the geometry ofthe deflected member, we consider triangle ABB�. Assuming that anglesare small, we can write

(4)

Substituting uA given by Equation 3 into Equation 4 gives

(5)

Alternatively, we can compute identical values of uA and �B by com-puting the shear and moment produced by the angle change u applied asan elastic load to the beam at point B (see Fig. 9.20a). Summing momentsabout support C to compute RA produces

(6)

After RA is computed, we draw the shear and moment curves in the usualmanner (see Fig. 9.20b and c). Since the shear just to the right of supportA equals RA, we observe that the shear given by Equation 6 is equal to theslope given by Equation 3. Further, because the shear is constant between

RA �u 1L x 2

L

u 1L x 2 RAL � 0

A �MC � 0

¢B �u 1L x 2x

L

¢B � uAx

uA �u 1L x 2

L

uA �¢C

L

¢C � u 1L x 2

L – xx

x

L

L

RA RC

�

�

(a)

Shear (slope)

(b)

Moment (deflection)

(c)

(L – x)L

�

(L – x)xMB = �B =L

�

CBA

Figure 9.20: (a) Angle change u applied as aload at point B; (b) shear produced by load uequals slope in real beam; (c) moment producedby u equals deflection in real beam (see Fig.9.19).

the support and point B, the slope of the real structure must also be con-stant in the same region.

Recognizing that the moment MB at point B equals the area under theshear curve between A and B, we find

(7)

Comparing the value of deflections at B given by Equations 5 and 7, weverify that the moment MB produced by load u is equal to the value of �B

based on the geometry of the bent beam. We also observe that the maxi-mum deflection occurs at the section where the shear produced by theelastic load is zero.

Sign Convention

If we treat positive values of the M>EI diagram applied to the beam as adistributed load acting upward and negative values of M>EI as a down-ward load, positive shear denotes a positive slope and negative shear anegative slope (see Fig. 9.21). Further, negative values of moment indi-cate a downward deflection and positive values of moment an upwarddeflection.

Examples 9.13 and 9.14 illustrate the use of the elastic load methodto compute deflections of simply supported beams.

Compute the maximum deflection and the slope at each support for thebeam in Figure 9.22a. Note that EI is a constant.

SolutionAs shown in Figure 9.22b, the M>EI diagram is applied to the beam asan upward load. The resultants of the triangular distributed loads betweenAB and BC, which equal 720>EI and 360>EI, respectively, are shown byheavy arrows. That is,

and

Using the resultants, we compute the reactions at supports A and C. Theshear and moment curves, drawn in the conventional manner, are plottedin Figure 9.22c and d. To establish the point of maximum deflection, we

1

2 16 2 a 120

EIb �

360

EI

1

2 112 2 a 120

EIb �

720

EI

¢B � MB �u 1L x 2x

L


MEI

y

x

(a)

Figure 9.21: (a) Positive elastic load; (b) posi-tive shear and positive slope; (c) positive momentand positive (upward) deflection.

y

+V +V V�+ =

M�+ =

x

(b)

(c)

+M +M

E X A M P L E 9 . 1 3

Section 9.4 Elastic Load Method 321

locate the point of zero shear by determining the area under the loadcurve (shown shaded) required to balance the left reaction of 480>EI.

(1)

Using similar triangles (see Fig. 9.22b) yields

and (2)

Substituting Equation 2 into Equation 1 and solving for x give

To evaluate the maximum deflection, we compute the moment at x � 9.8ft by summing moments of the forces acting on the free body to the left ofa section through the beam at that point. (See shaded area in Fig. 9.22b.)

Using Equation 2 to express y in terms of x and substituting x � 9.8 ft,we compute

The values of the end slopes, read directly from the shear curve in Fig-ure 9.22c, are

Compute the deflection at point B of the beam in Figure 9.23a. Also locatethe point of maximum deflection; E is a constant, but I varies as shown onthe figure.

SolutionTo establish the M>EI curve, we divide the ordinates of the moment curve(see Fig. 9.23b) by 2EI between A and B and by EI between B and C. Theresulting M>EI diagram is applied to the beam as an upward load in

uA � 480

EI uC �

600

EI

¢max � 3135.3

EI T

¢max � M � 480

EI19.8 2

1

2 xy a x

3b

x � 296 � 9.8 ft

y �10

EI x

y

120> 1EI 2 �x

12

1

2 xy �

480

EI

A CB

10 kips

30 kips

20 kips(a)

12� 6�

y

Elastic loads(b)

Shear (slope)(c)

Moment (deflection)(d)

8�

720EI

x

360EI

120EI

480EI

480EI

–

3135.3EI

–

600EI

600EI

6� 4�

x = 9.8�

Figure 9.22: (a) Beam; (b) beam loaded by M>EIdiagram; (c) variation of slope; (d) deflected shape.

E X A M P L E 9 . 1 4


Figure 9.23c. The maximum deflection occurs 4.85 m to the left of sup-port C, where the elastic shear equals zero (Fig. 9.23d ).

To compute the deflection at B, we compute the moment produced atthat point by the elastic loads using the free body shown in Figure 9.23e.Summing moments of the applied loads about B, we compute


MB

R1 =

(a)

(b)

(c)

(d)

(e)

6 m

x = 2 m

4 m3.5 m1 m

0.5 m

4.85 m

6 m3 m1003

kN

300 kN •m

Moment(kN •m)

Elasticloads

Shear (slope)

300

583.33EI

100EI

150EI

200EI

75EI

R2 =300EI

R3 =600EI

391.67EI

391.67EI

583.33EI

–

1003

kN

A B

I2I

C

391.67EI

600EI

Figure 9.23


Section 9.5 Conjugate Beam Method 323

9.5 Conjugate Beam Method

In Section 9.4 we used the elastic load method to compute slopes anddeflections at points in a simply supported beam. The conjugate beammethod, the topic of this section, permits us to extend the elastic loadmethod to beams with other types of supports and boundary conditionsby replacing the actual supports with conjugate supports to produce aconjugate beam. The effect of these fictitious supports is to imposeboundary conditions which ensure that the shear and moment, producedin a beam loaded by the M>EI diagram, are equal to the slope and thedeflection, respectively, in the real beam.

To explain the method, we consider the relationship between the shearand moment (produced by the elastic loads) and the deflected shape ofthe cantilever beam shown in Figure 9.24a. The M>EI curve associatedwith the concentrated load P acting on the real structure establishes thecurvature at all points along the axis of the beam (see Fig. 9.24b). Forexample, at B, where the moment is zero, the curvature is zero. On the otherhand, at A the curvature is greatest and equal to –PL>EI. Since the curva-ture is negative at all sections along the axis of the member, the beam isbent concave downward over its entire length, as shown by the curvelabeled 1 in Figure 9.24c. Although the deflected shape given by curve 1is consistent with the M>EI diagram, we recognize that it does not repre-sent the correct deflected shape of the cantilever because the slope at theleft end is not consistent with the boundary conditions imposed by thefixed support at A; that is, the slope (and the deflection) at A must bezero, as shown by the curve labeled 2.

Therefore, we can reason that if the slope and deflection at A must bezero, the values of elastic shear and elastic moment at A must also equalzero. Since the only boundary condition that satisfies this requirement isa free end, we must imagine that the support A is removed—if no supportexists, no reactions can develop. By establishing the correct slope anddeflection at the end of the member, we ensure that the member is ori-ented correctly.

On the other hand, since both slope and deflection can exist at thefree end of the actual cantilever, a support that has a capacity for shearand moment must be provided at B. Therefore, in the conjugate beam wemust introduce an imaginary fixed support at B. Figure 9.24d shows theconjugate beam loaded by the M>EI diagram. The reactions at B in the

¢B � 1150

EI T

¢B � MB �600

EI 12 2

391.67

EI 16 2

AB

P

(a)

–

(b)

(c)

(d)

1

2�A A�= = 0

L

PLEI

A

A

B

B

–PLEI

PL2

2EIRB =

PL3

3EIMB =

MEI

Figure 9.24: (a) Deflected shape of a cantileverbeam. (b) M>EI diagram which establishes varia-tion of curvature. (c) Curve 1 shows a deflectedshape consistent with M>EI diagram in (b) but notwith the boundary conditions at A. Curve 2 showscurve 1 rotated clockwise as a rigid body until theslope at A is horizontal. (d ) Conjugate beam withelastic load.

conjugate beam produced by the elastic load [M>EI diagram] give theslope and deflection in the real beam.

Figure 9.25 shows the conjugate supports that correspond to a vari-ety of standard supports. Two supports that we have not discussed previ-ously—the interior roller and the hinge—are shown in Figure 9.25d ande. Since an interior roller (Fig. 9.25d ) provides vertical restraint only, thedeflection at the roller is zero but the member is free to rotate. Becausethe member is continuous, the slope is the same on each side of the joint.To satisfy these geometric requirements, the conjugate support must have


Real Support Conjugate Support

(a)

(b)

Pin or roller= 0�≠ 0�

(c)

(d)

(e)

Pin or roller= 0≠ 0

MV

Free end≠ 0�≠ 0�

Fixed end≠ 0≠ 0

M

M

V

Fixed end= 0�= 0�

L�R�

Free end= 0= 0

MV

Interior support= 0�≠ 0L =� R� ≠ 0VL = VR

Hinge= 0M

Hinge≠ 0�may have

different values different valuesand

R�L�

R�L� may haveVL

VL

and VR

VR

Interior roller≠ 0M

M

VL VR

V

V

Figure 9.25: Conjugate supports.


zero capacity for moment (thus, zero deflection), but must permit equalvalues of shear to exist on each side of the support—hence the hinge.

Since a hinge provides no restraint against deflection or rotation in areal structure (see Fig. 9.25e), the device introduced into the conjugatestructure must ensure that moment as well as different values of shear oneach side of the joint can develop. These conditions are supplied by usingan interior roller in the conjugate structure. Moment can develop becausethe beam is continuous over the support, and the shear obviously canhave different values on each side of the roller.

Figure 9.26 shows the conjugate structures that correspond to eightexamples of real structures. If the real structure is indeterminate, the con-jugate structure will be unstable (see Fig. 9.26e to h). You do not have tobe concerned about this condition because you will find that the M>EIdiagram produced by the forces acting on the real structure produceselastic loads that hold the conjugate structure in equilibrium. For exam-ple, in Figure 9.27b we show the conjugate structure of a fixed-end beamloaded by the M>EI diagram associated with a concentrated load appliedat midspan to the real beam. Applying the equations to the entire struc-ture, we can verify that the conjugate structure is in equilibrium with respectto both a summation of forces in the vertical direction and a summationof moments about any point.

actual beam conjugate beam

(a)

(b)

(c)

(d)

(e)

( f )

(g)

(h)

Figure 9.26: Examples of conjugate beams.

PL2

(a)

(b)

L

PL8EI

L4

– PL8EI

– PL8EI

L4

Figure 9.27: (a) Fixed-ended beam with con-centrated load at midspan; (b) conjugate beamloaded with M>EI curve. The conjugate beam,which has no supports, is held in equilibrium bythe applied loads.

In summary, to compute deflections in any type of beam by the con-jugate beam method, we proceed as follows.

1. Establish the moment curve for the real structure.2. Produce the M>EI curve by dividing all ordinates by EI. Variation of

E or I may be taken into account in this step.3. Establish the conjugate beam by replacing actual supports or hinges

with the corresponding conjugate supports shown in Figure 9.25.4. Apply the M>EI diagram to the conjugate structure as the load, and

compute the shear and moment at those points where either slope ordeflection is required.

Examples 9.15 to 9.17 illustrate the conjugate beam method.

For the beam in Figure 9.28 use the conjugate beam method to determinethe maximum value of deflection between supports A and C and at thetip of the cantilever. EI is constant.

SolutionThe conjugate beam with the M>EI diagram applied as an upward load isshown in Figure 9.28c. (See Fig. 9.25 for the correspondence betweenreal and conjugate supports.) Compute the reaction at A by summingmoments about the hinge.

RA �480

EI

18RA 720 110 2

EI

360 14 2EI

� 0

A �Mhinge � 0


E X A M P L E 9 . 1 5

10 kips 20 kips

30 kips

(a)

(b)

(c)

= RA

= RD

x

y

18�

120EI

120

Moment(kip •ft)

6�

3600EI

480EI

600EI

12� 6� 6�

A B C D

Figure 9.28: (a) Beam details; (b) moment curve;(c) conjugate beam with elastic loads; (d) elasticshear (slope); (e) elastic moment (deflection).

(d)

(e)

9.8�

6�

480EI

–

3136EI

–

600EI

3600EI

straight

600EI


Compute RD.

Draw the shear and moment curves (see Fig. 9.28d and e). Moment at D(equals area under shear curve between C and D) is

Locate the point of zero shear to the right of support A to establish the loca-tion of the maximum deflection by determining the area (shown shaded)under the load curve required to balance RA.

(1)

From similar triangles (see Fig. 9.28c),

(2)

Substituting Equation 2 into Equation 1 and solving for x give

Compute the maximum value of negative moment. Since the shear curveto the right of support A is parabolic, area � bh.

Compute the deflection at D.

Determine the maximum deflection of the beam in Figure 9.29a. We knowthat EI is a constant.

SolutionThe ordinates of the moment diagram produced by the concentrated loadsacting on the real structure in Figure 9.29a are divided by EI and applied

¢D � MD �3600

EI

¢max � Mmax �2

319.8 2 a

480

EIb �

3136

EI

23

x � 296 � 9.8 ft

y120EI

�x

12 and y �

10

EI x

1

2 xy �

480

EI

MD �600

EI 16 2 �

3600

EI

RD �600

EI

720

EI

360

EI

480

EI RD � 0

c �Fy � 0

E X A M P L E 9 . 1 6


as a distributed load to the conjugate beam in Figure 9.29b. We next dividethe distributed load into triangular areas and compute the resultant (shownby heavy arrows) of each area.

Compute RE.

RE �81P

EI

36P

EI 16 2

18P

EI 14 2

18P

EI 18 2

54P

EI 110 2 12RE � 0

B �MC � 0


P 2P

MA = 18P 2P

P

(a)

(b)

(c)

(d)

Shear = slope

Moment = deflection

6�

4� 4� 6�

10�

18PEI

54PEI

18PEI

36PEI

90PEI

–

756PEI

–

45PEI

81PEI

RE =

81PEI

135PEI

RC =

18PEI

6PEI

6PEI

6� 6� 6�

EC

A B C D E

Figure 9.29: (a) Beam; (b) conjugate structureloaded by M>EI diagram; (c) slope; (d) deflection.



Compute RC.

To establish the variation of slope and deflection along the axis of thebeam, we construct the shear and moment diagrams for the conjugatebeam (see Fig. 9.29c and d ). The maximum deflection, which occurs atpoint C (the location of the real hinge), equals 756P>EI. This value isestablished by evaluating the moment produced by the forces acting onthe conjugate beam to the left of a section through C (see Fig. 9.29b).

Compare the magnitude of the moment required to produced a unit valueof rotation (uA � 1 rad) at the left end of the beams in Figure 9.30a andc. Except for the supports at the right end—a pin versus a fixed end—thedimensions and properties of both beams are identical, and EI is con-stant. Analysis indicates that a clockwise moment M applied at the leftend of the beam in Figure 9.30c produces a clockwise moment of M>2 atthe fixed support.

SolutionThe conjugate beam for the pin-ended beam in Figure 9.30a is shown inFigure 9.30b. Since the applied moment M� produces a clockwise rota-tion of 1 rad at A, the reaction at the left support equals 1. Because theslope at A is negative, the reaction acts downward.

To compute the reaction at B, we sum moments about support A.

RB �M¿L6EI

0 � RBL M¿L2EI

aL3b

A �MA � 0

RC �135P

EI

54P

EI

18P

EI

18P

EI

81P

EI

36P

EI RC � 0

c �Fy � 0

E X A M P L E 9 . 1 7


Summing forces in the y direction, we express M� in terms of the prop-erties of the member as

(1)

The conjugate beam for the fixed-end beam in Figure 9.30c is shownin Figure 9.30d. The M>EI diagram for each end moment is drawn sepa-rately. To express M in terms of the properties of the beam, we sum forcesin the y direction.

(2)

NOTE. The absolute flexural stiffness of a beam can be defined as thevalue of end moment required to rotate the end of a beam—supported ona roller at one end and fixed at the other end (see Fig. 9.30c)—throughan angle of 1 radian. Although the choice of boundary conditions is some-what arbitrary, this particular set of boundary conditions is convenientbecause it is similar to the end conditions of beams that are analyzed bymoment distribution—a technique for analyzing indeterminate beams andframes covered in Chapter 13. The stiffer the beam, the larger the momentrequired to produce a unit rotation.

If a pin support is substituted for a fixed support as shown in Figure9.30a, the flexural stiffness of the beam reduces because the roller doesnot apply a restraining moment to the end of the member. As this exam-ple shows by comparing the values of moment required to produce a unitrotation (see Eqs. 1 and 2), the flexural stiffness of a pin-ended beam isthree-fourths that of a fixed-end beam.

M¿ �3

4 M

M¿M

�3EI>L4EI>L

M �4EI

L

0 � 1 ML

2EI

1

2 ML

2EI

c �Fy � 0

M¿ �3EI

L

0 � 1 M¿L2EI

M¿L6EI

c �Fy � 0


A�

(a)

= 1 rad

A

L

M�

M

B

(b)

(c)

(d)

L3

RA = 1

x =

1

M�EI

M�L2EI

MEI

ML2EI

ML2EIML

4EI

M�L6EI

M2

RB =

A� = 1 rad

Figure 9.30: Effect of end restraint on flexuralstiffness. (a) Beam loaded at A with far end pinned;(b) conjugate structure for beam in (a) loadedwith M>EI diagram; (c) beam loaded at A with farend fixed; (d ) conjugate structure for beam in (c)loaded with M>EI diagram.


Section 9.6 Design Aids for Beams 331

9.6 Design Aids for Beams

To be designed properly, beams must have adequate stiffness as well asstrength. Under service loads, deflections must be limited so that attachednonstructural elements—partitions, pipes, plaster ceilings, and windows—will not be damaged or rendered inoperative by large deflections. Obvi-ously floor beams that sag excessively or vibrate as live loads are appliedare not satisfactory. To limit deflections under live load, most buildingcodes specify a maximum value of live load deflection as a fraction ofthe span length—a limit between 1/360 to 1/240 of the span length iscommon.

If steel beams sag excessively under dead load, they may be cambered.That is, they are fabricated with initial curvature by either rolling or byheat treatment so that the center of the beam is raised an amount equal toor greater than the dead load deflection (Fig. 9.31). Example 10.12 illus-trates a simple procedure to relate curvature to camber. To camber rein-forced concrete beams, the center of the forms may be raised an amountequal to or slightly greater than the dead load deflections.

In practice, designers usually make use of tables, in handbooks anddesign manuals to evaluate deflections of beams for a variety of loadingand support conditions. The Manual of Steel Construction published bythe American Institute of Steel Construction (AISC) is an excellentsource of information.

Table 9.1 gives values of maximum deflections as well as momentdiagrams for a number of support and loading conditions of beams. Wewill make use of these equations in Example 9.18.

A simply supported steel beam spanning 30 ft carries a uniform dead loadof 0.4 kip/ft that includes the weight of the beam and a portion of the floorand ceiling supported directly on the beam (Fig. 9.32). The beam is alsoloaded at its third points by two equal concentrated loads that consist of14.4 kips of dead load and 8.2 kips of live load. To support these loads,the designer selects a 16-in-deep steel wide-flange beam with a modulusof elasticity E � 30,000 ksi and a moment of inertia I � 758 in4.

(a) Specify the required camber of the beam to compensate for thetotal dead load deflection and 50 percent of the live load deflection.

(b) Verify that under live load only, the beam does not deflect morethan 1/360 of its span length. (This provision ensures the beam will notbe excessively flexible and vibrate when the live load acts.)

SolutionWe first compute the required camber for dead load, using equations fordeflection given by cases 1 and 3 in Table 9.1.

camber

Figure 9.31: Beam fabricated with camber.

E X A M P L E 9 . 1 8

wD = 0.4 kip/ft

PL = 8.2 kipsPD = 14.4 kips

PL = 8.2 kipsPD = 14.4 kips

10� 10� 10�

Figure 9.32: Beam, connected to columns by clipangles attached to web, is analyzed as a simplysupported determinate beam.


332

L

w

�MAX

�MAX =

wL2

M

M

MM

M

wL2

1 5

2 6

3 7

4 8

wL2

8

5wL4

384EI

L

P

�MAX

�MAX =

PL4

L2

L2

P2

P2

PL3

48EI

L

P P

�MAX

�MAX = (3L2 – 4a2)

Pa

a a

PP

Pa24EI

L

P

P�MAX

�MAX = PL3

3EI

M = – PL

L

�MAX

�MAX (L + a)=

M

–Pa

a

P 1 +PaL

aL

Pa2

3EI

L

w

P

�MAX

�MAX =–wL4

384EI

wL2

12

wL2

12– wL2

12

wL2

24

wL2

12

�MAX PL8

PL

P2

P2

PL8

PL8– PL

8–

PL8

L

P

�MAX = PL3

192EI

L

wL�MAX

�MAX = wL4

8EI

– = MwL2

2

wL2

2

w

wL2

wL2

TABLE 9.1

Moment Diagrams and Equations for Maximum Deflection

Summary 333

(a) Dead load deflection produced by uniform load is

Dead load deflection produced by concentrated loads is

(b) Allowable live load deflection is

Therefore, it is OK.

Summary

• The maximum deflections of beams and frames must be checked toensure that structures are not excessively flexible. Large deflectionsof beams and frames can produce cracking of attached nonstructuralelements (masonry and tile walls, windows, and so forth) as well asexcessive vibrations of floor and bridge decks under moving loads.

• The deflection of a beam or frame is a function of the bendingmoment M and the member’s flexural stiffness, which is related to amember’s moment of inertia I and modulus of elasticity E. Deflectionsdue to shear are typically neglected unless members are very deep,shear stresses are high, and the shear modulus G is low.

• To establish equations for the slope and deflection of the elastic curve(the deflected shape of the beam’s centerline), we begin the study ofdeflections by integrating the differential equation of the elastic curve

This method becomes cumbersome when loads vary in a complexmanner.

d 2y

dx 2 �M

EI

L

360�

30 � 12

360� 1 in 7 0.6 in

Required camber � ¢DT 1

2¢L � 1.37

0.60

2� 1.67 in

¢L � 0.6 in

Total dead load deflection, ¢DT � ¢D1 ¢D2 � 0.32 1.05 � 1.37 in

¢D2 � 1.05 in

¢D2 �Pa 13L2 4a2 2

24EI�

14.4 110 2 33 130 2 2 4 110 2 2 4 11728 224 130,000 2 1758 2

¢D1 �5wL4

384EI�

5 10.4 2 130 2 4 11728 2384 130,000 2 1758 2 � 0.32 in

Live load deflection, ¢L �Pa 13L2 4a2 2

24EI�

8.2 1102 33 1302 2 4 1102 2 4 11728224 130,000 2 1758 2


• Next we consider the moment-area method, which utilizes the M>EIdiagram as a load to compute slopes and deflections at selectedpoints along the beam’s axis. This method, described in Section 9.3,requires an accurate sketch of the deflected shape.

• The elastic load method (a variation of the moment-area method),which can be used to compute slopes and deflections in simplysupported beams, is reviewed. In this method, the M>EI diagram isapplied as a load. The shear at any point is the slope, and the momentis the deflections. Points of maximum deflections occur where theshear is zero.

• The conjugate beam method, a variation of the elastic load method,applies to members with a variety of boundary conditions. Thismethod requires that actual supports be replaced by fictitious supportsto impose boundary conditions that ensure that the values of shearand moment in the conjugate beam, loaded by the M>EI diagram,are equal at each point to the slope and deflection, respectively, ofthe real beam.

• Once equations for evaluating maximum deflections are establishedfor a particular beam and loading, tables available in structuralengineering reference books (see Table 9.1) supply all the importantdata required to analyze and design beams.


PROBLEMSSolve Problems P9.1 to P9.6 by the double integrationmethod. EI is constant for all beams.

P9.1. Derive the equations for slope and deflection forthe cantilever beam in Figure P9.1. Compute the slopeand deflection at B. Express answer in terms of EI.

P9.2. Derive the equations for slope and deflection forthe beam in Figure P9.2. Compare the deflection at Bwith the deflection at midspan.

P9.3. Establish the equations for slope and deflectionfor the beam in Figure P9.3. Evaluate the slope at eachend of the beam.

P9.4. Derive the equations for slope and deflection forthe beam in Figure P9.4. Locate the point of maximumdeflection and compute its magnitude.

AB

P

L

EI = constant

P9.1

L

AB

w

P9.2

L

AM

B

P9.3

L

A

w

B

P9.4

Problems 335

P9.5. Establish the equations for slope and deflectionfor the beam in Figure P9.5. Evaluate the magnitude ofthe slope at each support. Express answer in terms of EI.

P9.6. Derive the equations for slope and deflection forthe beam in Figure P9.6. Determine the slope at eachsupport and the value of the deflection at midspan. (Hint:Take advantage of symmetry; slope is zero at midspan.)

Solve Problems P9.7 to P9.28 by the moment-areamethod. Unless noted otherwise, EI is a constant for allmembers. Answers may be expressed in terms of EIunless otherwise noted.

P9.7. Compute the slope and deflection at points B andC in Figure P9.7.

P9.8. (a) Compute the slope at A and C and the deflec-tion at B in Figure P9.8. (b) Locate and compute themagnitude of the maximum deflection.

P9.9. Compute the slope at A and the deflection atmidspan for the beam in Figure P9.9.

P9.10. (a) Compute the slope at A and the deflection atmidspan in Figure P9.10. (b) If the deflection at midspanis not to exceed 1.2 in, what is the minimum requiredvalue of I? E � 29,000 kips/in2.

P9.11. (a) Find the slope and deflection at A in FigureP9.11. (b) Determine the location and the magnitude ofthe maximum deflection in span BC.

L

AM

M2

B

P9.5

P

L2

A B

L2

P9.6

3 m 2 m

A CB

6 kN/m

P9.7

P = 15 kips

12� 6�

A B C

P9.8

A

M M

CB

L2

L2

P9.9

A B

2II I

C D E

P = 30 kips

18�6� 6�

P9.10

AB

2II I

C D

10 kips

12�6� 6�

P9.11

P9.12. Compute the slopes of the beam in Figure P9.12on each side of the hinge at B, the deflection of the hinge,and the maximum deflection in span BCD. The elas-tomeric support at D acts as a roller.

P9.13. Compute the slope at A and the deflection atmidspan in Figure P9.13. Treat rocker at D as a roller.

P9.14. Determine the maximum deflection in span ABand the deflection of C in Figure P9.14.

P9.15. Determine the slope and deflection of point C inFigure P9.15. Hint: Draw moment curves by parts.

P9.16. Compute the deflection of A and the slope at Bin Figure P9.16. Hint: Plot moment curves by parts.Express answer in terms of EI.

P9.17. Determine the deflection of the hinge at C andthe slope at D in Figure P9.17. EI is constant.

P9.18. Compute the deflection of points B and D in Fig-ure P9.18. The elastomeric pad at C acts as a roller.


P P

L3

A

rocker

B DC

L3

L3

P9.13

L L3

AM

B C

P9.14

A BC

3 m 2 m

9 kN/m

30 kN

P9.15

P = 4 kips

w = 2 kips/ft

AB

C

12�6�

P9.16

A B DC

16 kips

6�24� 6�6�

P9.17

12 kips

elastomericpad

6� 6�

4�

CA

B

D

P9.18

A B

2I I

DC

P = 30 kips

9�9�12�

P9.12

Problems 337

P9.19. Compute the vertical and horizontal componentsof the deflection at E and the vertical deflection of A inFigure P9.19.

P9.20. The moment of inertia of the girder in FigureP9.20 is twice that of the column. If the vertical deflectionat D is not to exceed 1 in and if the horizontal deflectionat C is not to exceed 0.5 in, what is the minimum requiredvalue of the moment of inertia? E � 29,000 kips/in2. Theelastomeric pad at B is equivalent to a roller.

P9.21. Compute the vertical displacement of the hingeat E in Figure P9.21.

P9.22. Determine the horizontal and vertical deflectionof D produced by the 6-kip load at C in Figure P9.22.

P9.23. Compute the slope at A and the horizontal andvertical components of deflection at point D in FigureP9.23.

P9.24. What value of force P is required at B in FigureP9.24 if the vertical deflection at B is to be zero?

5 kips

6�

6�

15� 4�

CBA

D

E

P9.19

B C D

2I2I

I

A

4 kips

6� 12�

9�

P9.20

B E C

P = 9 kipsP = 9 kips

DA

12� 12�

6�

6�

P9.21

6 kips

6�

18�

12�

A

DB

C

P9.22

B CD

A

9 kips

6 kips

6�12�

6�

P9.23

6� 4�

AB

w = 3 kips/ft

P = ?

P9.24

P9.25. What value of force F is required at midspan ifthe beam in Figure P9.25 is not to deflect vertically atthat point? Express your answer in terms of w and L.

P9.26. Compute the horizontal displacement of joint Bin Figure P9.26. The moment diagram produced by the12-kip load is given. The base of the columns at pointsA and E may be treated as fixed supports. Hint: Begin bysketching the deflected shape, using the moment dia-grams to establish the curvature of members. Momentsin units of kip�ft.

P9.27. The frame shown in Figure P9.27 is loaded byequal but opposite loads at A and B. Compute the rota-tion at E and the vertical deflection at B. Given: E � 200GPa, IAB � 800 � 106 mm4, and ICD � 400 � 106 mm4.

P9.28. The frame shown in Figure P9.28 is loaded by ahorizontal load at B. Compute the horizontal displace-ments at B and D. For all members E � 200 GPa and I �500 � 106 mm4.

Solve Problems P9.29 to P9.33 by the conjugate beammethod.

P9.29. Compute the slope and deflection at point C ofthe cantilever beam in Figure P9.29. EI is constant.


A

B C D

E

12 kips

58.1

58.1

66.4

44.8

44.8

19.1 32.3

15� 30�

15�

P9.26

200 kN200 kN

A E B

D

C

4 m 4 m

2 m

2 m

P9.27

A

B

C

D

100 kN

12 m

5 m

5 m

P9.28

10 kips

4�6�

A B C

P9.29

P = ?L

A CB

w

L

P9.25

Problems 339

P9.30. Compute the deflection at points A and C (atmidspan) of the beam in Figure P9.30. Also I varies asshown on sketch.

P9.31. Compute the slope and deflection at point C andthe maximum deflection between A and B for the beamin Figure P9.31. The reactions are given, and EI is con-stant. The elastomeric pad at B is equivalent to a roller.

P9.32. Determine the flexural stiffness of the beam inFigure P9.32 (see Example 9.17 for criteria) for (a)moment applied at A and (b) moment applied at C. E isconstant.

P9.33. Using the conjugate beam method, compute themaximum deflection in span BD of the beam in FigureP9.33 and the slope on each side of the hinge.

P9.34. Solve Problem P9.11 by the conjugate beammethod.



P9.37. For the beam shown in Figure P9.37, use theconjugate beam method to compute the vertical deflec-tion and the rotation to the left and right of C. Given: E �200 GPa, IAC � 2I, and ICF � I, where I � 50 � 106

mm4.

AB

2I

CL

I I

C DE

30 kips 30 kips

12�6� 6�

P9.30

6 kips

9 kips

3 kips

3�9�

AMA = 9 kip •ft

BC

P9.31

A B C

L2

I 2I

L2

P9.32

A B D EC

8 kips

6� 6�

hinge

6� 4�

P9.33

10 m5 m5 m

10 kN

100 kN •m

5 m5 m

FEDCBA

P9.37

Practical Applications of DeflectionComputations

P9.38. The reinforced concrete girder shown in FigureP9.38a is prestressed by a steel cable that induces acompression force of 450 kips with an eccentricity of 7in. The external effect of the prestressing is to apply anaxial force of 450 kips and equal end moment MP �262.5 kip�ft at the ends of the girder (Fig. P9.38b). Theaxial force causes the beam to shorten but produces nobending deflections. The end moments MP bend thebeam upward (Fig. P9.38c) so that the entire weight of

the beam is supported at the ends, and the member actsas a simply supported beam. As the beam bends upward,the weight of the beam acts as a uniform load to producedownward deflection. Determine the initial camber ofthe beam at midspan immediately after the cable is ten-sioned. Note: Over time the initial deflection will increasedue to creep by a factor of approximately 100 to 200percent. The deflection at midspan due to the two endmoments equals ML2>(8EI) Given: I � 46,656 in4, A �432 in2, beam weight wG � 0.45 kip/ft, and E � 5000kips/in2.


450kips

450kips

60�

A

ASection A-A

prestressedtendon

36�

initial camber

7�

12�

450 kips450 kips

MP = 262.5 kip •ft MP = 262.5 kip •ft

–262.5 kip •ft

wG

Beam dimensions(a)

Forces applied to concrete by prestress(b)

moment diagram

Deflected shape; prestress and weight of beam act(c)

P9.38

Problems 341

P9.39. Because of poor foundation conditions, a 30-in-deep steel beam with a cantilever is used to support anexterior building column that carries a dead load of 400kips and a live load of 100 kips (Fig. P9.39). What is themagnitude of the initial camber that should be inducedat point C, the tip of the cantilever, to eliminate the

deflection produced by the total load? Neglect thebeam’s weight. Given: I � 46,656 in4 and ES � 30,000ksi. See case 5 in Table 9.1 for the deflection equation.The clip angle connection at A may be treated as a pinand the cap plate support at B as a roller.

A B C

PD = 400 kipsPL = 100 kips

camber?scale notcorrect

9�24�

P9.39

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